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Question 1 Arrange the steps in the order that will allow you to determine the concentration of the base solution + Write a balance equation for the reaction between the analyte and the titrant + Calc.
Question 1: Arrange the steps in the order that will allow you to determine the concentration of the base solution: + Write a balance equation for the reaction between the analyte and the titrant + Calculate the number of moles of titrant using the volume of titrant required and the concentration of titrant + Calculate the number of moles of analyte using the stoichiometric coefficients of the equation + Calculate the concentration of the analyte using the number of moles of analyte and the volume of analyte titrated Question 2: Write the balance titration equation: Fe3+ + Cu+ Fe2+ + Cu2+ Complete the two half-reaction that occur at the Pt indicator electrode Write the half-reactions as reductions: Half reaction of Copper : Cu2+ + e- Cu+ (Estd = 0.161V) Half reaction of Iron : Fe3+ + e- Fe2+ (Estd = 0.767V) Select the two equations given in the pictures below that can be used to determine the cell voltage at different points in the titration E of the Ag/AgCl electrode is 0.197V Equation 1: E = ECu2+/Cu+ - 0.0592 log – ESCE = 0.161 – 0.0592 log – 0.197 (V) Equation 2: E = EFe3+/Fe2+ - 0.0592 log – ESCE = 0.767 – 0.0592 log – 0.197 (V) Calculate the value of E for the cell after each of the given volume of the 1.50 ml Cu++ titrant have been added: Initial moles of Fe3+ = 0.02 x 100 ml = mmol Initial moles of Cu+ = 0.1 x 1.5 ml = 0.15 mmol After reaction: [Fe2+] = [Fe3+] = Ecell = 0.767 – 0.0592 log – ESCE = 0.767 – 0.0592 log – 0.197 = 0.635 V Question 3: Electricity is the flow of electrons The question relate to how electricity is quantified Electron are charged particles The amount of charge that passes per unit time is called current The driving force for the electron (i.e., the reason they are flowing in the first place is measured by potential(thế năng) Charge is measured in coulombs (C) Current is measured in amperes (A) Question 4: A solution of Cu+ is undergoing a redox titration by the addition of Ce4+ The standard reduction half-reactions and potentials are given: Select a suitable indicator to observe the end point: Which color change would you observe the end point? Cu+ + Ce4+ Cu2+ + Ce3+ The two half cell: Ce4+ +e- Ce3+ Estd = 1.72V Cu2+ + e- Cu+ Estd = 0.161V Cu+ Cu2+ + e- (Anode reaction) Ce4+ + e- Ce3+ (Cathode reaction) Ecell = Eanode – Ecathode = 1.72 – 0.161 = 1.559V The suitable indicator is 5,6 – dimethylphenanthroline The color turns from yellow-green to red Question 5: Match each of the electrical circuit terms with their definition Electrical potential: The work that can be done to move an electric charge from one point to another (volts) Electrical charge(E): the amount of positive or negative particles (coulombs) Current: the quantity of charge moving through a circuit each second (amperes) Question 6: End point: The point in the titration when the added amount of standard reagent is equal to the amount of analyte being titrated Equivalence point: The point in the titration when a change in the analyte solution is observed, indicating equivalency Analyte: A solution of unknown concentration Indicator: It is added to the analyte solution and aids in the observation of the completion of the reaction Titrant: The standard reagent of known concentration that is added from a buret to the analyte solution Molarity: A common concentration unit – moles per liter Titration: A volumetric analysis technique used to determine unknown concentration of a solution by using the known concentration of other Question 7: Making the table about the indicator and their pH Analyzing the curve and observe what pH range that the pH changes rapidly Equivalence point Compare it with the table Suitable indicator Question 8: Which of the factors would have a major effect (greater than 0.1 pH units) on the pH range for the end point of an acid-base titration + Titrant concentration + Strength of acid or base Question 9: When a weak acid (HA) is titrated with a strong base, such as NaOH what species are present in the weak acid solution before the titration is started ? HA, H+ (H3O+) , A- , H2O Question 10: Dilute solution with a same weak acid (pKa = 4.5) ranging in concentration from x 10^-6 to x 10^-2 M (0.002 to 20mM) are titrated with a strong base that is five times more concentrated than the acid At low concentration of a weak acid, how does the concentration of the acid affect the shape of the titration curve? Answer : The lower the concentration of the acid, the sharper the inflection at the equivalence point More information: + The shape depends on the pKa and the concentration of acid Question 11: Consider a monoprotic weak acid (HA) that is titrated with a strong base What is the relationship between the strength of the weak acid and the pH of the solution at the equivalence point? Answer: The weaker the acid, the higher the pH at the equivalence point Explanation: According to Bronsted lowry concept : Stronger the acid , weaker will be it's conjugate base Or Weaker the acid, stronger will be it's conjugate base Question 12: Identify the equilibrium shift that takes place after each reaction is cooled in a refrigerator in which direction does the equilibrium of this reaction shift due to cooling? 2H2S + SO2 Answer: ∆H is given in negative form So, it is an exothermic reaction With decrease in temperature, equilibrium shifts in product side according to Lechateliers principle Toward the products ∆H is negative The right shift is exothermic an exothermic reaction So : Increasing the temperature Shift to left Toward the reactants Decreasing the temperature Shift to right Toward the products Question 13: A particular gas-phase reaction has an equilibrium constant of Kp = 0.30 A mixture is prepared where all the reactants and products are in their standard states Which direction will the reaction proceed? We know that Kp = If Kp > The reaction is favored It will proceed toward products that is right Oppositely, it will proceed toward reactants that is left Qp > Kp (0.3) The reaction will proceed to the left Question 14: In the Fajans titration of Hg2(2+) , NaCl is added to produce the precipitate Hg2Cl2 The end point of the titration is detected with bromophenol blue What charge you expect the precipitate to have after the equivalence point ? Answer: Before equivalence mercurous ion will be in the solution So formed precipitate of mercurous chloride will adsorb excess of mercurous ion so solution will be positively charged In Fazan's methid, Before the equivalance point of titration, the precipitate is negatively charged since solution consists of excess Cl- ions which are adsorbed ob the precipitate, giving negative charge to precipitate But after the equivalance point of titration, Hg2+ ions are in excess, so the surface of precipitate becomes positively charged with adsorbed layer of excess Hg 2+ ions Question 15: Methyl orange is an indicator used for detecting the pH of acidic solutions The red, protonated form of the indicator (Hln) dissociates into a yellow, deprotonated form (In-) as shown in the reaction The pKa for methyl orange is 3.4 Hln(aq) Red Yellow Identifying the color of a solution containing methyl orange indicator at pH The color will be red due to the acidic solution(pH = 2) If pH = 3.4 pKa Equivalence point Orange If pH = Basic solution Yellow Question 16+17: Both liquid bleach, ClO- and iodate,IO3- , can react with iodide ions, IIn a reaction with iodide ions, liquid bleach is best classified as ? Reaction of ClO- with IClO- + I- > Cl- + IOThis reaction occur in presence of OH- as catalyst Iis oxidized to IO- and ClO- is reduced to Cl- Liquiđ bleach is best classified as oxidizing agent Reaction of IO3- with IIO3- + I- + 6H+ -> I2 + H2O IO3- is reduced to I2 while I- is oxidized to I2 Iodate ion is best classified as oxidizing agent Question 18: Equal volumes of two different weak acids are titrated with 0.20M NaOH, resulting in the following titration curves Which curve corresponds to the titration of the more concentrated weak acid solution? To determine the curve of more and less concentrated weak acid we need to look at titration curves carefully When the titration is of more concentrated weak acid with strong base the pH at its equivalence point is equal to while in less concentrated weak acid and strong base titration pH at equivalence point will be greater than (as base is strong) In the blue curve it is clearly visible that pH at equivalence point is nearly equal to while in red curve pH is greater than 7(between - 10) So we can say blue curve corresponds to titration of more concentrated weak acid solution Question 19: Cell membranes contain channels that allow ions to cross the phospholipid bilayer A particular K+ channel carries a current of 1.61 pA How many K+ ions, N, pass through the channel n 1.65 m/s ? Hint : Current is the rate of charge flow Consider how much charge a single K+ ion possesses Charge on K+ ion = 1.6 x 10^-19 C Total charge passed = Q = It = (1.61 x 10^-12) x (1.65 x 10^-3) = 2.6565 x 10^-15 C Number of K+ ion = N = Q/e = Question 20: Simply living, an 87.5 kg human being will consume approximately 20.0 mol of O2 per day To provide energy for the human, the O2 is reduced to H2O during food oxidation by the reaction O2 + 4H+ + 4ea) Determine the current (I) generated by the human per day In this case, the current is defined as the flow of electrons (e-) to O2, from the food the human consumes b) Reduction of O2 can occur by a donation of electrons from nicotinamide adenine dinucleotide (NADH) When this occurs, the electron experience a potential drop of 1.00 V Determine the power output(P) from the 87.5kg human ANSWER a) O2 + 4H+ + 4e20 80 Moles of electron = n(e) = 20 x = 80 moles Q(charge) = n(e) x F(Faraday constant) = 80 x 96485 = 7718800 C = nNF Current(I) = = 89.3 A b) Power(P) = U(potential drop) x I (Current) = x 89.3 = 89.3 W Question 21: During the titration of Cl- with Ag+, the precipitate is in equilibrium with the ion pair AgCl(aq) Determine the concentration of AgCl(aq) during the precipitation titration The formation constant for the ion pair is 4.9 x 10^-4 and the Ksp for AgCl is 1.8 x 10^-10 Ksp = [Ag+][Cl-] = x*x = 1.8 x 10^-10 X = 1.34 x 10^-5 = [Ag+] = [Cl-] Ag+ + Cl- AgCl Kf = [AgCl] = 8.82 x 10^-14 M Question 22: A 2.108g sample of a solid mixture containing only potassium carbonate (MM = 138.2058 g/mol) and potassium bicarbonate (MM = 100.1154 g/mol) is dissolved in distilled water A volume of 35.79 mL of a 0.744 M HCl standard solution is required to titrate the mixture to a bromocresol green end point Calculate the weight percent of potassium carbonate and potassium bicarbonate in the mixture Moles of HCl = 0.02663 moles K2CO3 + 2HCl 2KCl + CO2 + H2O X 2x KHCO3 + HCl KCl + CO2 + H2O Y y x 2x + y = 0.02663 138.2058x + 100.1154y = 2.108 x = 0.009 ; y = 0.00863 Mass of K2CO3 = 1.244g Mass of KHCO3 = 0.864g Weight percent of K2CO3 = Weight percent of KHCO3 = 100% – 59.01% = 40.99% Question 23: a) Au + 3HNO3 + 4HCl HAuCl4 + 3NO2 + 3H2O b) In this reaction, the oxidation state of Cl- does not change It does not participate in the redox process(oxidation-reduction) Hence, HCl supplies chloride ions to form a complex ion with the oxidized gold Question 24: You perform an electrochemical reaction in which 0.800 mol of Cu+ are reduced to solid Cu a) How many coulombs of charge are transferred ? b) How many electrons are in this amount of charge ? a) Cu+ Cu + e0.8 0.8 Q = n x F = 0.8 x 96486 = 77188 C b) Moles of electrons = 0.8 x 6.022 x 10^-23 = 4.8176 x 10^-23 Number of electrons in this amount of charge = Q/n(e) = 1.602 x 10^27 Question 25: What is the concentration of a 50.8 mL solution of HBr that is completely titrated by 29.00 ml of a 0.200 M NaOH solution ? Moles of NaOH = 0.0058 moles Moles of HBr = Moles of NaOH = 0.0058 moles The concentration of a 50.8 mL solution of HBr = 0.114 M Question 26: Balance the redox reaction by inserting the appropriate coefficients Fe3+ + NO2- Fe2+ + NO3- Adding water : Fe3+ + NO2- + H2O Fe2+ + NO3Adding H+ : 2Fe3+ + NO2- + H2O 2Fe2+ + NO3- + 2H+ This is balanced chemical equation in acidic medium Question 27: Enter the solubility product expression for Al(OH)3 (s) Al3+ + 3OH- Al(OH)3 The solubility product = Ksp = [Al3+] Question 28: Enter the solubility expression for Mg3(PO4)2 3Mg2+ + 2PO43- Mg3(PO4)2 Ksp = ... n(e) = 20 x = 80 moles Q(charge) = n(e) x F(Faraday constant) = 80 x 96485 = 7718800 C = nNF Current(I) = = 89.3 A b) Power(P) = U(potential drop) x I (Current) = x 89.3 = 89.3 W Question 21:... expression for Al(OH)3 (s) Al3+ + 3OH- Al(OH)3 The solubility product = Ksp = [Al3+] Question 28: Enter the solubility expression for Mg3(PO 4)2 3Mg2+ + 2PO43- Mg3(PO 4)2 Ksp = ... dinucleotide (NADH) When this occurs, the electron experience a potential drop of 1.00 V Determine the power output(P) from the 87.5kg human ANSWER a) O2 + 4H+ + 4e20 80 Moles of electron = n(e) = 20 x